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(A) In the steady state,the rate of heat flow through both cylinders is the same.Let $r_1$ be the radius of the smaller cylinder and $r_2$ be the radi

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) In the steady state,the rate of heat flow through both cylinders is the same.Let $r_1$ be the radius of the smaller cylinder and $r_2$ be the radius of the larger cylinder.Given $r_2 = 2 r_1$,the cross-sectional areas are $A_1 = \pi r_1^2$ and $A_2 = \pi r_2^2 = \pi (2 r_1)^2 = 4 \pi r_1^2 = 4 A_1$.The rate of heat flow is given by $\frac{dQ}{dt} = \frac{K A \Delta T}{L}$.For the first cylinder: $\frac{dQ}{dt} = \frac{K_1 A_1 (300 - 200)}{L} = \frac{K_1 A_1 (100)}{L}$.For the second cylinder: $\frac{dQ}{dt} = \frac{K_2 A_2 (200 - 100)}{L} = \frac{K_2 (4 A_1) (100)}{L}$.Equating the two rates of heat flow:$\frac{K_1 A_1 (100)}{L} = \frac{K_2 (4 A_1) (100)}{L}$.$K_1 = 4 K_2$.Therefore,$\frac{K_1}{K_2} = 4$.

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